Q116 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Mathematics
$\lim _{n \rightarrow \infty}\left[\begin{array}{l}\frac{1}{n}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3} \\ +\ldots+\frac{1}{125 n}\end{array}\right]=$
- A$\frac{3}{8}$
- B$\frac{15}{32}$
- Correct answerC$\frac{12}{25}$
- D$\frac{35}{72}$
Solution
Given,
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3}+\ldots \frac{1}{125 n}\right] \\
& =\lim _{n \rightarrow \infty}\left[\begin{array}{l}
\frac{n^2}{(n+0)^3}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3} \\
+\ldots+\frac{n^2}{(n+4 n)^3}
\end{array}\right] \\
& =\lim _{n \rightarrow \infty} \sum_{r=0}^{4 n} \frac{n^2}{(n+r)^3}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Let } \frac{r}{n}=x \text { and } \frac{1}{n}=d x=\int_0^4 \frac{d x}{(1+x)^3}=\left[\frac{(1+x)^{-2}}{-2}\right]_0^4 \\
& =-\frac{1}{2}\left[\frac{1}{5^2}-1\right]=\frac{1}{2}\left(1-\frac{1}{25}\right)=\frac{12}{25} .
\end{aligned}
$$
Hence, option (c) is correct.
Hence, the correct answer is option: 'C'.